Finding solutions to x2- b2=b2-1 is a bit too easy these days, so let b itself be the multiple of a square, say, yz2
For both y>1 and z>1, prove that y is not a square, or find a counter-example.
(In reply to
Oinking solution. by Jer)
I found the first 3 values of b (5, 29, 169)
Jer, b=(yz^2). 5 and 29 are not numbers in this form, since they are prime and cannot be factored into a square and another factor, still less squared factors such as (az)^2.
169 is, since 169=(1^2*13^2) and indeed gives a solution for x=239. Then we have (yz^2) = (1^2*13^2) or (yz^2) = (13^2*1^2) with both y and z square; but either y=1 or z=1, contrary to the second stipulation.
You are right that x^2-(25y)^2=(25y)^2-1 has a solution, and indeed recurrent solutions, since equivalently x^2 - 1250y^2 =-1, see e.g. here: http://www.alpertron.com.ar/QUAD.HTM
But can y itself ever be square in this, or indeed any other equation of the form x^2-(yz^2)^2=(yz^2)^2-1, if both y and z are greater than 1? That is the question.
Edited on September 26, 2012, 2:03 am
|
|
Posted by broll
on 2012-09-26 00:42:04 |
re: Oinking solution.
